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Graphs Isomorphic to Their Path Graphs

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Graphs Isomorphic to Their Path Graphs 127 (2002) MATHEMATICA BOHEMICA No. 3, 473–480 GRAPHS ISOMORPHIC TO THEIR PATH GRAPHS 1 2 Martin Knor ,Bratislava,Ľudovít Niepel ,Kuwait (Received November 15, 2000) Abstract. We prove that for every number n 1, the n-iterated P3-path graph of G is isomorphic to G if and only if G is a collection of cycles, each of length at least 4. Hence, G is isomorphic to P3(G) if and only if G is a collection of cycles, each of length at least 4. ∼ Moreover, for k 4 we reduce the problem of characterizing graphs G such that Pk(G) = G to graphs without cycles of length exceeding k. Keywords: line graph, path graph, cycles MSC 2000 : 05C38 1. Introduction Let G be a graph, k 1, and let Pk be the set of paths of length k in G.The vertex set of a path graph Pk(G)isthesetPk. Two vertices of Pk(G)arejoinedby an edge if and only if the edges in the intersection of the corresponding paths form a path of length k − 1inG, and their union forms either a cycle or a path of length k + 1. It means that the vertices are adjacent if and only if one can be obtained from the other by “shifting” the corresponding paths in G. Path graphs were investigated by Broersma and Hoede in [2] as a natural general- ization of line graphs, since P1(G) is the line graph L(G)ofG (for further connections to line graphs see [6]). Traversability of P2-path graphs is studied in [9], and a char- acterization of P2-path graphs is given in [2] and [7]. Distance properties of path graphs are studied in [1], [3] and [5], and in [4] graphs with connected P3-path graphs are characterized. When a new function on graphs appears, one of the very first problems is to determine the fixed points of the function, i.e., graphs that are isomorphic to their 1 Supported by VEGA grant 1/6293/99. 2 Supported by Kuwait University grant #SM 02/00. 473 images. It is well known (and trivial to prove) that a connected graph G is isomorphic to its line graph L(G) if and only if G is a cycle. In [2] it is proved that a connected graph G is isomorphic to its P2-path graph if and only if G is a cycle. We remark that Pk(G) is not necessarily a connected graph if G is connected and k 2. However, slight modifications of the proof in [2] give rise to the following theorem: Theorem A. Let G be a graph isomorphic to P2(G). Then each component of G is a cycle. i However, even stronger theorem follows from [5]. Let Pk(G)denotethei-iterated i i−1 0 Pk-path graph of G, i.e., Pk(G)=Pk(Pk (G)) if i>0, and Pk (G)=G.Wehave Theorem B. Let G be a graph and n anumber,n 1, such that G is isomorphic n to P2 (G). Then each component of G is a cycle. By now, only a little is known about Pk-path graphs for k 3.In[8]LiandZhao proved the following theorem: Theorem C. Let G be a connected graph isomorphic to P3(G).ThenG is a cycle of length greater than or equal to 4. In this paper we generalize Theorem C to an analogue of Theorem A for P3-path graphs: Theorem 1. Let G be a graph isomorphic to P3(G). Then each component of G is a cycle of length greater than or equal to 4. In fact, we prove more. We prove an analogue of Theorem B for P3-path graphs: Theorem 2. Let G be a graph and n anumber,n 1, such that G is isomorphic n to P3 (G). Then each component of G is a cycle of length greater than or equal to 4. We remark that the proof of Theorem C in [8] is analogous to the proof of Theo- rem A (in a weaker form) in [2], and since it is based on some counting arguments, it is not clear how to extend it to the proof of Theorem 2. In fact, our approach to the problem is completely different. At present, we are not able to generalize Theorem 1 to Pk-path graphs for k 4. The problem seems to be too complicated in general; for a solution for graphs in which every component contains a “large” cycle see Lemma 3. Thus, for k 4it may be useful to start with trees, see also Corollary 6. However, even for trees the problem appears to be hard. In this connection we pose the following 474 i Problem. Does there exist a tree T and a number k, k 4, such that Pk(T ) is a nonempty forest for every i 0? P G k G If such a tree does not exist, then k(G) for every number and a forest . (We remark that the problem is trivial for k =1;fork = 2 it is solved negatively in [5]; and for k = 3 it is solved negatively by Lemma 7.) In the next two sections we prove Theorem 2. In Section 2 we present some general results for Pk-path graphs when k 2,andSection3isdevotedtoP3-path graphs of trees. 2. General results We use standard graph-theoretic notation. Let G be a graph. The vertex set and theedgesetofG, respectively, are denoted by V (G)andE(G). If v is a vertex of G then degG(v) denotes the degree of v in G. For two subgraphs, H1 and H2 of G,we denote by H1 ∪ H2 the union of H1 and H2 in G,andbyH1 ∩ H2 their intersection. A path and a cycle, respectively, of length l are denoted by Pl and Cl. For easier handling of paths of length k in G (i.e., the vertices of Pk(G)) we make the following agreement. We denote the vertices of Pk(G) (as well as the vertices of G) by small letters a, b,..., while the corresponding paths of length k in G are denoted by capital letters A, B,....ItmeansthatifA is a path of length k in G and a is a vertex in Pk(G), then a must be the vertex corresponding to the path A. Throughout this section, the symbol k is used for the length of paths producing the path graph. I.e., we consider here only Pk-path graphs, k 2. By a large cycle we mean a cycle of length greater than k. A cycle whose length does not exceed k is a small cycle. n ∼ Lemma 3. Let G be a graph such that Pk (G) = G for some n 1 and k 2. Then every component of G containing a large cycle is isomorphic to a single cycle. C P C ∼ n C ∼ ÈÖÓ Ó.If is a large cycle then k( ) = C, and hence Pk ( ) = C,too.As n ∼ n Pk (G) = G, all large cycles in Pk (G) are images of large cycles in G. If a and b are adjacent vertices in Pk(G), then A and B share a path of length k − 1. This implies that if C is a large cycle in G,thenPk(C) does not contain a n n chord in Pk(G), and hence, Pk (C) does not contain a chord in Pk (G), either. Since n the number of large cycles in G is equal to the number of large cycles in Pk (G), no large cycle contains a chord in G. Suppose that there is a large cycle in G with a vertex incident to an edge outside this cycle. For every large cycle C,letIG(C) denote the total number of edges outside C that are incident to a vertex of C.AsC does not contain a chord, 475 n n I Pk C · IG C I n P C · IG C I G Pk (G)( ( )) = 2 ( ), and Pk (G)( k ( )) = 2 ( ). Let ( )denotethe n n maximum value of IG(C), where C is a large cycle in G.ThenI(Pk (G)) = 2 ·I(G). n Since I(G) > 0, Pk (G) is not isomorphic to G. Lemma 4. Every small cycle in Pk(G) hasanevenlength. C a ,a ,...,a l P G l k ÈÖÓ Ó.Let =( 1 2 l) be a small cycle of length in k( ), .If u and v are adjacent vertices in Pk(G), then U and V share a path of length k − 1. Since l k, A1,A2,...,Al share a path P of length t k − (l − 1) 1inG.Let P =(p0,p1,...,pt). Assume that A1 =(a1,0,a1,1,...,a1,k),...,Al =(al,0,al,1,...,al,k) are denoted so that for every i,1 i l,wehavei0 <it,whereai,i0 = p0 and ai,it = pt.Then i0 and (i +1)0 have different parity, 1 i<l.As(a1,a2,...,al)formsacycle,10 and l0 have different parity, too, and hence l is even. We remark that Lemma 3 and Lemma 4 reduce the examination of graphs G for n ∼ which Pk (G) = G, to graphs without large cycles and without odd small cycles. Hence, to complete the proof of Theorem 2 it remains to study forests. K K Definition. Atree t is obtained from a claw 1,3 subdividing each edge of 1,3 by t − 1 vertices, see Figure 1 for 3.Atree t1,t2 is obtained from two paths of length 2t1 central vertices of which are joined by a path of length t2,seeFigure2 for 1,2 and Figure 3 for 2,1.
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